Soit la médiane et soit la moyenne d'un échantillon aléatoire de taille d'une distribution . Comment puis-je calculer ?
Intuitivement, en raison de l'hypothèse de normalité, il est logique de prétendre que et c'est effectivement la bonne réponse. Cela peut-il être montré avec rigueur?
Ma pensée initiale était d'aborder ce problème en utilisant la distribution normale conditionnelle qui est généralement un résultat connu. Le problème est que, comme je ne connais pas la valeur attendue et par conséquent la variance de la médiane, je devrais calculer ceux qui utilisent la statistique er ordre. Mais c'est très compliqué et je préfère ne pas y aller à moins d'y être absolument obligé.
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normal-distribution
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JohnK
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Réponses:
Laissez désignent l'échantillon original et Z le vecteur aléatoire avec des entrées Z k = X k - ˉ X . Alors Z est centré normal (mais ses entrées ne sont pas indépendantes, comme le montre le fait que leur somme est nulle avec une probabilité totale). En tant que fonction linéaire de X , le vecteur ( Z , ˉ X ) est normal donc le calcul de sa matrice de covariance suffit à montrer que Z est indépendant de ˉX Z Zk=Xk−X¯ Z X (Z,X¯) Z .X¯
En ce qui concerne , on voit que Y = ˉ X + T où T est la médiane des Z . En particulier, T ne dépend que de Z donc T est indépendant de ˉ X , et la distribution deY Y=X¯+T T Z T Z T X¯ Z is symmetric hence T is centered.
Enfin,
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The sample median is an order statistic and has a non-normal distribution, so the joint finite-sample distribution of sample median and sample mean (which has a normal distribution) would not be bivariate normal. Resorting to approximations, asymptotically the following holds (see my answer here):
with
whereX¯n is the sample mean and μ the population mean, Yn is the sample median and v the population median, f() is the probability density of the random variables involved and σ2 is the variance.
So approximately for large samples, their joint distribution is bivariate normal, so we have that
whereρ is the correlation coefficient.
Manipulating the asymptotic distribution to become the approximate large-sample joint distribution of sample mean and sample median (and not of the standardized quantities), we have
So
We have that2f(v)=2/σ2π−−√ due to the symmetry of the normal density so we arrive at
where we have usedv=μ . Now the standardized variable is a standard normal, so its absolute value is a half-normal distribution with expected value equal to 2/π−−−√ (since the underlying variance is unity). So
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The answer isx¯ .
Letx=(x1,x2,…,xn) have a multivariate distribution F for which all the marginals are symmetric about a common value μ . (It does not matter whether they are independent or even are identically distributed.) Define x¯ to be the arithmetic mean of the xi, x¯=(x1+x2+⋯+xn)/n and write x−x¯=(x1−x¯,x2−x¯,…,xn−x¯) for the vector of residuals. The symmetry assumption on F implies the distribution of x−x¯ is symmetric about 0 ; that is, when E⊂Rn is any event,
Applying the generalized result at /stats//a/83887 shows that the median ofx−x¯ has a symmetric distribution about 0 . Assuming its expectation exists (which is certainly the case when the marginal distributions of the xi are Normal), that expectation has to be 0 (because the symmetry implies it equals its own negative).
Now since subtracting the same valuex¯ from each of a set of values does not change their order, Y (the median of the xi ) equals x¯ plus the median of x−x¯ . Consequently its expectation conditional on x¯ equals the expectation of x−x¯ conditional on x¯ , plus E(x¯ | x¯) . The latter obviously is x¯ whereas the former is 0 because the unconditional expectation is 0 . Their sum is x¯, QED.
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This is simpler than the above answers make it. The sample mean is a complete and sufficient statistic (when the variance is known, but our results do not depend on the variance, hence will be valid also in the situation when the variance is unknown). Then the Rao-Blackwell together with the Lehmann-Scheffe theorems (see wikipedia ...) will imply that the conditional expectation of the median, given the arithmetic mean, is the unique minimum variance unbiased estimator of the expectationμ . But we know that is the arithmetic mean, hence the result follows.
We did also use that the median is an unbiased estimator, which follows from symmetry.
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